Last week I posted this puzzle…
Albert works in a clock factory. He has noticed that his favourite clock in the factory takes 7seconds to strike 7 o’clock. How long will the same clock take to strike 10 o’clock?
Update: I have had loads of emails about this arguing that it is impossible to solve! So, here is the new question ‘Albert thinks that they answer will definitely be 10 seconds. Is he right?’ 10 points for anyone who gets the right answer to that, and 100 points for anyone who tries to work out an answer to the original question!
If you didn’t try to solve it, have a go now. For everyone else, the answer is after the break (although I have a funny feeling this one is going to cause a bit of debate!)
The original solution was as follows. When the clock chimes 7 o’clock it will have 7 strikes and six pauses:
I_I_I_I_I_I_I
We know that this took 7 seconds, so each pause will take 7/6 seconds.
Chiming 10 o’clock involves 10 chimes and 9 pauses
I_I_I_I_I_I_I_I_I_I
And thus this will take 9*7/6 seconds, which works out at 10.5 seconds.
However, this does not take into account the actual length of the chimes, which gives rise to a far more complicated situation. It is very unlikely that the answer would be 10 seconds, so Albert is wrong to think that, but what ideas did you have about how long it might take if you take the chimes into account?
First time I have added an update from my iPhone ! The three hour solution is great. So, not only does this have a lateral solution but the mathematical one is counter-intuitive!
June 14, 2009 at 11:03 pm |
Let x be the length of a chime and y be the length of a pause. Then 7 chimes in 6 seconds means that 7x+6y=7. If it were also the case that we had ten chimes in ten seconds, then 10x+9y=10. Using basic linear algebra, this system of equations has the unique solution (x=1, y=0). The fact that this is a solution is perhaps not surprising, but the significant fact is that this solution is unique, so unless the pause takes absolutely no time at all, there is no way that it’d take ten seconds for ten chimes.
Additionally, the matrix
7 6
10 9
is invertible, so there is a unique solution (x,y) for any amount of time you’d like ten chimes to take (although some won’t work physically since x or y would be negative).
June 14, 2009 at 11:21 pm |
Oh my god. I actually got it.
June 14, 2009 at 11:47 pm |
It’s 10.5 seconds if you consider the end to be the start of the last chime. It’s 11.6 seconds if you consider the end to be the start of where the 11th chime would begin. And if you consider the end to be the total amount of time it takes for the chime to fade, of course we don’t have enough information.
June 15, 2009 at 8:57 am
That’s not correct. If you consider the end to be where the next chime would begin, then the original 7 chimes taking 7 seconds would have been exactly 1 second per “chime-until-start-of-next”, and the answer would be 10.
June 15, 2009 at 2:04 am |
My answer was three hours.
June 15, 2009 at 4:54 am
I was trying to solve this puzzle and was unwilling to succumb to doing any math when my mother walked up behind me and after a moment said, “I think it’s three hours.” Brilliant.
Glad there’s more who came up with this solution too!
June 15, 2009 at 2:19 am |
Sally, you are awesome! I love your solution.
June 15, 2009 at 2:27 am |
Oh wait, no I didn’t. I took each chime and then each pause as a seperate unit. I knew there had to be 6 pauses and 7 chimes in 7 seconds, and that makes 13 units total. I divided the 7 seconds by that and got 0.538. 10 chimes meant 9 pauses, so I multiplied 19 by around 0.538 to get 10.23.
But seriously, that’s a fraction of a second off. I think I still deserve a cookie.
Oh yeah, I decided the chimes and pauses were of equal length because nobody indicated otherwise, and I wasn’t going to earn 100 points by sitting around the entire weekend making up arbitrary chime and pause lengths.
June 15, 2009 at 3:09 am |
Sally,
I thought the same thing!
Cheers.
June 15, 2009 at 3:26 am |
I thought of it entirely differently, I thought it would take 3 hours and 3 seconds from the end of the seven o’clock chime to the end of the ten o’clock chime.
June 15, 2009 at 3:32 am |
Got it right (of course)
The 0.1 nanosecond guy
June 15, 2009 at 3:45 am |
Miko’s already solved it, but my 2c worth of further simplification: if the length of the chime is x seconds, the total time to strike 10 chimes will be 10.5 – 0.5x, e.g. if the chime is 0.2s long, the total time will be 10.4s. The limits for the total time are 10s (where x=1 and y=0, i.e. no interval between the chimes) and 10.5s (where x=0s and y=7/6s, i.e. instantaneous chimes).
x cannot be greater than 1, or else it violates the initial 7x+6y=7 equation given by Miko above; if x>1 then y<0, which, unless the clock also happens to be a TARDIS, is slightly impossible. And of course x can't be less than 0s, again unless the clock also happens to be one of the aforementioned TARDISs (TARDII?).
June 15, 2009 at 5:26 am |
Can Albert check time on a more precise scale than 1 second?
I am hard pressed to think so.
So whatever the mathematical answer might be, Albert *will* find the 10 o’clock chimes to be 10 seconds.
Btw. I did get the formula. But the funny thing is, the longer a chime reverbs and thus gets closer to a full second, the closer the pause gets to 0 and the closer the answer gets to 10 seconds.
June 15, 2009 at 5:27 am |
@myself… the closer the answer gets to 11 seconds…. oops.
June 15, 2009 at 5:29 am |
Awww, what the heck, I decided to spring for a digital clock for Albert… what’s his address?
June 15, 2009 at 5:55 am |
I took the 7 seconds to include the full 4 lines of the westminster quarters chimes, plus the 7 strikes of the hour…
http://en.wikipedia.org/wiki/Westminster_Quarters
So my by my calculations, the majority of the 7 seconds is used for the Westminster Quarters full hour chime (about 6 seconds 1.5 seconds per verse) and the last 1 second for the 7 chimes of the hour, so the difference for an additional 3 chimes is about 3/7ths of a second, or negligible.
June 15, 2009 at 6:45 am |
100% right … As I said, just like the barcode problem!
June 15, 2009 at 6:58 am |
This is too far fetched IMHO.
June 15, 2009 at 6:59 am |
I like the first posters simultaneous equation approach – I saw both cases as possible answers but without realizing the continum of possible answers depending on the pause, just the two extreme cases. I now see that the strike can happen at any point while the last chime is still ringing too such that the last chime ending ringing could take a different time too – so:
7x + y = 7 and
10x + y = 10
(would only be true if y=0 so if a bell doesn’t keep ringing after being struck, which is not true) – so if the end of the chimes is taken to be when the last stike stops ringing, then with the pauses less then the time a chime rings 10secs would be wrong too.
Disappointed it is a trivial maths ‘gotcha’ rather than a psychologically revealing puzzle though, but pleased I saw the fairly obvious answer anyhow
June 15, 2009 at 7:29 am |
I got the answer three hours and three and a half seconds, which is the length of time from the seventh chime at seven o’clock and the tenth at ten o’clock.
I suspect that the chiming mechanisms of most clocks do not take into account the length of the chime’s ring itself. My personal interpretation here is that a clock finishes striking a time when the last bell is “struck”, not when that bell finishes sounding. It’s the attack, baby, not the decay.
June 15, 2009 at 8:10 am |
I did not see that one at all. I feel ashamed.
Good puzzle
June 15, 2009 at 8:26 am |
My first solution was thinking there must be a twist so I thought the answer must be timed from the end of the seventh chime of seven o’clock and the end of the tenth chime of 10 o’clock! This answer assumed we were talking about 7am and 10am. So I came up with the same 3 hours 3 and a half seconds as Lafayette just said.
My second solution was the 10.5 seconds!
Then I thought we could look at it being 7am and 10pm but of course that would be the second time it would strike 10 o’clock so I realised that wasn’t a viable answer but if it had of been it would have been 15 hours 3 and a half seconds.
June 15, 2009 at 9:05 am |
There has been no mention of an “intro” yet either.
Many clocks have a small chimed piece of music before ringing the hour.
This could further complicate the issue.
7x + 6y = 7 – z
June 15, 2009 at 9:08 am |
I had three hours answer as my solution to the first problem but once you reworded the question I got the other answer.
I thought the first answer was a lateral thinking problem that I have come to expect from these! The second was more of a straight maths puzzle.
June 15, 2009 at 9:31 am |
I trust that the gentle reader will deem to excuse any slips o’the keyboard, but here is my quick analysis of the conundrum:
There are two essentially believable scenarios* in this perfect puzzle world, apart from the tricky notions to which I alluded in a previous post.
(More on that later)
Scenario:
a) Where the period in question is measured from the start of the first chime to the end of the last chime.
b) Where the period in question is measured from the start of the first chime to the beginning of the last chime.
These two scenarios resolve into 4 variables:
1) The number of chimes (Which I shall call ‘n’)
2) The duration of the chime (Which I shall call ‘Tc’)
3) The duration of the intervening silence (Which I shall call ‘Ts’)
4) The target ‘total’ duration. (Which I shall call ‘Tt’)
For scenario a), the equation is:
Tt=n*Tc+(n-1)*Ts
(e.g.: #_#_#_#_#_#_#, for n=7)
For scenario b), the equation is:
Tt=n*Tc+(n-0)*Ts
=n*Tc+n*Ts
=n*(Tc+Ts)
(e.g.: #_#_#_#_#_#_, for n=7)
__________________
Lets work the a) case for both 7 & 10:
n=7?
a) Tt=n*Tc+(n-1)*Ts
=7*Tc+6*Ts
n=10?
a) Tt=n*Tc+(n-1)*Ts
=10*Tc+9*Ts
—————
—————
Now the b) case for both 7 & 10:
n=7?
b) Tt=n*(Tc+Ts)
=7*(Tc+Ts)
n=10?
b) Tt=n*(Tc+Ts)
=10*(Tc+Ts)
—————
Now, the question is: are n=7 & n=10 produce and equal (Total Period)/n in either of these scenarios?
This amounts to asking does:
a1)
Tt(7)/n==Tt(10)/n??
or
(7*Tc+6*Ts)/7==(10*Tc+9*Ts)/10 ?
or
b1)
7*(Tc+Ts)/7==10*(Tc+Ts)/10 ?
(this obviously simplifies to: Tc+Ts==Tc+Ts, which is trivially true.)
—————–
a1) Is true, provided that the silence lasts for 0 seconds, otherwise false.
b1) Is ALWAYS true!
So, I was guardedly correct** (No) for scenario a), but quite wrong for scenario b)!!
The resolution of this puzzle has therefore been rigorously proven to depend upon the assumption of how the total chime length is determined.
For start of chime 1, to END of chime ‘n’, the answer is NEGATIVE.
For start of chime 1, to START of chime ‘n’, the answer is POSITIVE!*
———————
OK: The tricky notions, regarding the question as implying the temporal distance between the 7 o’clock chimes, and the 10 o’clock chimes.
There are several options available here for consideration:
The gaps:
7am to 10am ~ 3h
7am to 10pm ~ 3+12=15h
7pm to 10pm ~ 3+12=15h
7pm to 10pm ~ 3h
Which resolve down to two distinct durations:
3h & 15h
Both of which are subject to the a) & b) condition of the judgement of the length of a “strike”.
These trick answers are bogus, in my opinion, but don’t tell Richard that I said that, or he may join the BCA!
But do inform him (for that is, I understand, the prime target of these puzzles), that all of the above machinations had occurred to me in my head, (not to commit in writing), in under say, a minute.
Possibly closer to 30 seconds than a full minute.
It is a burden to be a systems analyst with Asperger’s!
_____________
* That the chime durations are equal, and the silence durations are equal.
** In assuming that the silent gaps were not zero length.
June 15, 2009 at 9:32 am |
I trust that the gentle reader will deem to excuse any slips o’the keyboard, but here is my quick analysis of the conundrum:
There are two essentially believable scenarios* in this perfect puzzle world, apart from the tricky notions to which I alluded in a previous post.
(More on that later)
Scenario:
a) Where the period in question is measured from the start of the first chime to the end of the last chime.
b) Where the period in question is measured from the start of the first chime to the beginning of the last chime.
These two scenarios resolve into 4 variables:
1) The number of chimes (Which I shall call ‘n’)
2) The duration of the chime (Which I shall call ‘Tc’)
3) The duration of the intervening silence (Which I shall call ‘Ts’)
4) The target ‘total’ duration. (Which I shall call ‘Tt’)
For scenario a), the equation is:
Tt=n*Tc+(n-1)*Ts
(e.g.: #_#_#_#_#_#_#, for n=7)
For scenario b), the equation is:
Tt=n*Tc+(n-0)*Ts
=n*Tc+n*Ts
=n*(Tc+Ts)
(e.g.: #_#_#_#_#_#_, for n=7)
__________________
Lets work the a) case for both 7 & 10:
n=7?
a) Tt=n*Tc+(n-1)*Ts
=7*Tc+6*Ts
n=10?
a) Tt=n*Tc+(n-1)*Ts
=10*Tc+9*Ts
—————
—————
Now the b) case for both 7 & 10:
n=7?
b) Tt=n*(Tc+Ts)
=7*(Tc+Ts)
n=10?
b) Tt=n*(Tc+Ts)
=10*(Tc+Ts)
—————
Now, the question is: are n=7 & n=10 produce and equal (Total Period)/n in either of these scenarios?
This amounts to asking does:
a1)
Tt(7)/n==Tt(10)/n??
or
(7*Tc+6*Ts)/7==(10*Tc+9*Ts)/10 ?
or
b1)
7*(Tc+Ts)/7==10*(Tc+Ts)/10 ?
(this obviously simplifies to: Tc+Ts==Tc+Ts, which is trivially true.)
—————–
a1) Is true, provided that the silence lasts for 0 seconds, otherwise false.
b1) Is ALWAYS true!
So, I was guardedly correct** (No) for scenario a), but quite wrong for scenario b)!!
The resolution of this puzzle has therefore been rigorously proven to depend upon the assumption of how the total chime length is determined.
For start of chime 1, to END of chime ‘n’, the answer is NEGATIVE.
For start of chime 1, to START of chime ‘n’, the answer is POSITIVE!*
———————
OK: The tricky notions, regarding the question as implying the temporal distance between the 7 o’clock chimes, and the 10 o’clock chimes.
There are several options available here for consideration:
The gaps:
7am to 10am ~ 3h
7am to 10pm ~ 3+12=15h
7pm to 10pm ~ 3+12=15h
7pm to 10pm ~ 3h
Which resolve down to two distinct durations:
3h & 15h
Both of which are subject to the a) & b) condition of the judgement of the length of a “strike”.
These trick answers are bogus, in my opinion, but don’t tell Richard that I said that, or he may join the BCA!
But do inform him (for that is, I understand, the prime target of these puzzles), that all of the above machinations had occurred to me in my head, (not to commit in writing), in under say, a minute.
Possibly closer to 30 seconds than a full minute.
It is a burden to be a systems analyst with Asperger’s!
_____________
* That the chime durations are equal, and the silence durations are equal.
** In assuming that the silent gaps were not zero length.
June 15, 2009 at 9:35 am |
There are two essentially believable scenarios* in this perfect puzzle world, apart from the tricky notions to which I alluded in a previous post.
(More on that later)
Scenario:
a) Where the period in question is measured from the start of the first chime to the end of the last chime.
b) Where the period in question is measured from the start of the first chime to the beginning of the last chime.
These two scenarios resolve into 4 variables:
1) The number of chimes (Which I shall call ‘n’)
2) The duration of the chime (Which I shall call ‘Tc’)
3) The duration of the intervening silence (Which I shall call ‘Ts’)
4) The target ‘total’ duration. (Which I shall call ‘Tt’)
For scenario a), the equation is:
Tt=n*Tc+(n-1)*Ts
(e.g.: #_#_#_#_#_#_#, for n=7)
For scenario b), the equation is:
Tt=n*Tc+(n-0)*Ts
=n*Tc+n*Ts
=n*(Tc+Ts)
(e.g.: #_#_#_#_#_#_, for n=7)
__________________
Lets work the a) case for both 7 & 10:
n=7?
a) Tt=n*Tc+(n-1)*Ts
=7*Tc+6*Ts
n=10?
a) Tt=n*Tc+(n-1)*Ts
=10*Tc+9*Ts
—————
Now the b) case for both 7 & 10:
n=7?
b) Tt=n*(Tc+Ts)
=7*(Tc+Ts)
n=10?
b) Tt=n*(Tc+Ts)
=10*(Tc+Ts)
—————
Now, the question is: are n=7 & n=10 produce and equal (Total Period)/n in either of these scenarios?
This amounts to asking:
a1)
Tt(7)/n==Tt(10)/n??
or
(7*Tc+6*Ts)/7==(10*Tc+9*Ts)/10 ?
or
b1)
7*(Tc+Ts)/7==10*(Tc+Ts)/10 ?
(this obviously simplifies to: Tc+Ts==Tc+Ts, which is trivially true.)
—————–
a1) Is true, provided that the silence lasts for 0 seconds, otherwise false.
b1) Is ALWAYS true!
So, I was guardedly correct** (No) for scenario a), but quite wrong for scenario b)!!
The resolution of this puzzle has therefore been rigorously proven to depend upon the assumption of how the total chime length is determined.
For start of chime 1, to END of chime ‘n’, the answer is NEGATIVE.
For start of chime 1, to START of chime ‘n’, the answer is POSITIVE!*
———————
OK: The tricky notions, regarding the question as implying the temporal distance between the 7 o’clock chimes, and the 10 o’clock chimes.
There are several options available here for consideration:
The gaps:
7am to 10am ~ 3h
7am to 10pm ~ 3+12=15h
7pm to 10pm ~ 3+12=15h
7pm to 10pm ~ 3h
Which resolve down to two distinct durations:
3h & 15h
Both of which are subject to the a) & b) condition of the judgement of the length of a “strike”.
These trick answers are bogus, in my opinion, but don’t tell Richard that I said that, or he may join the BCA!
But do inform him (for that is, I understand, the prime target of these puzzles), that all of the above machinations had occurred to me in my head, (not to commit in writing), in under say, a minute.
Possibly closer to 30 seconds than a full minute.
It is a burden to be a systems analyst with Asperger’s!
_____________
* That the chime durations are equal, and the silence durations are equal.
** In assuming that the silent gaps were not zero length.
June 15, 2009 at 9:36 am |
Part 1?
There are two essentially believable scenarios* in this perfect puzzle world, apart from the tricky notions to which I alluded in a previous post.
(More on that later)
Scenario:
a) Where the period in question is measured from the start of the first chime to the end of the last chime.
b) Where the period in question is measured from the start of the first chime to the beginning of the last chime.
These two scenarios resolve into 4 variables:
1) The number of chimes (Which I shall call ‘n’)
2) The duration of the chime (Which I shall call ‘Tc’)
3) The duration of the intervening silence (Which I shall call ‘Ts’)
4) The target ‘total’ duration. (Which I shall call ‘Tt’)
For scenario a), the equation is:
Tt=n*Tc+(n-1)*Ts
(e.g.: #_#_#_#_#_#_#, for n=7)
For scenario b), the equation is:
Tt=n*Tc+(n-0)*Ts
=n*Tc+n*Ts
=n*(Tc+Ts)
(e.g.: #_#_#_#_#_#_, for n=7)
__________________
Lets work the a) case for both 7 & 10:
n=7?
a) Tt=n*Tc+(n-1)*Ts
=7*Tc+6*Ts
n=10?
a) Tt=n*Tc+(n-1)*Ts
=10*Tc+9*Ts
—————
Now the b) case for both 7 & 10:
n=7?
b) Tt=n*(Tc+Ts)
=7*(Tc+Ts)
n=10?
b) Tt=n*(Tc+Ts)
=10*(Tc+Ts)
—————
Now, the question is: are n=7 & n=10 produce and equal (Total Period)/n in either of these scenarios?
This amounts to asking:
a1)
Tt(7)/n==Tt(10)/n??
or
(7*Tc+6*Ts)/7==(10*Tc+9*Ts)/10 ?
or
b1)
7*(Tc+Ts)/7==10*(Tc+Ts)/10 ?
(this obviously simplifies to: Tc+Ts==Tc+Ts, which is trivially true.)
—————–
a1) Is true, provided that the silence lasts for 0 seconds, otherwise false.
b1) Is ALWAYS true!
So, I was guardedly correct** (No) for scenario a), but quite wrong for scenario b)!!
June 15, 2009 at 9:37 am |
I would say there are 6 chimes and 6 pauses (+intro) for the clock to ’strike’ 7
Once you’ve ’struck’ the 7th chime, we have struck 7 by definition, and do not have to include the amplitude envelope of the 7th chime. So each chime + pause takes 7/6 secs.
With no intro striking 10 would take 9*(7/6) = 10.5 secs
With intro there are 2 many variables 6c + i = 7, to solve c & i but
is it possible to strike 10 in 10 seconds given this scenario?: 9c + i = 10
Sure – each chime takes 1 second and there is a 1 second intro would solve.
I just got up, so the above may well be nonsense.
June 15, 2009 at 9:37 am |
There are two essentially believable scenarios* in this perfect puzzle world, apart from the tricky notions to which I alluded in a previous post.
(More on that later)
Scenario:
a) Where the period in question is measured from the start of the first chime to the end of the last chime.
b) Where the period in question is measured from the start of the first chime to the beginning of the last chime.
These two scenarios resolve into 4 variables:
1) The number of chimes (Which I shall call ‘n’)
2) The duration of the chime (Which I shall call ‘Tc’)
3) The duration of the intervening silence (Which I shall call ‘Ts’)
4) The target ‘total’ duration. (Which I shall call ‘Tt’)
Continued…
June 15, 2009 at 9:38 am |
…continued
For scenario a), the equation is:
Tt=n*Tc+(n-1)*Ts
(e.g.: #_#_#_#_#_#_#, for n=7)
For scenario b), the equation is:
Tt=n*Tc+(n-0)*Ts
=n*Tc+n*Ts
=n*(Tc+Ts)
(e.g.: #_#_#_#_#_#_, for n=7)
__________________
Lets work the a) case for both 7 & 10:
n=7?
a) Tt=n*Tc+(n-1)*Ts
=7*Tc+6*Ts
n=10?
a) Tt=n*Tc+(n-1)*Ts
=10*Tc+9*Ts
—————
Now the b) case for both 7 & 10:
n=7?
b) Tt=n*(Tc+Ts)
=7*(Tc+Ts)
n=10?
b) Tt=n*(Tc+Ts)
=10*(Tc+Ts)
—————
Now, the question is: are n=7 & n=10 produce and equal (Total Period)/n in either of these scenarios?
This amounts to asking:
a1)
Tt(7)/n==Tt(10)/n??
or
(7*Tc+6*Ts)/7==(10*Tc+9*Ts)/10 ?
or
b1)
7*(Tc+Ts)/7==10*(Tc+Ts)/10 ?
(this obviously simplifies to: Tc+Ts==Tc+Ts, which is trivially true.)
—————–
a1) Is true, provided that the silence lasts for 0 seconds, otherwise false.
b1) Is ALWAYS true!
So, I was guardedly correct** (No) for scenario a), but quite wrong for scenario b)!!
The resolution of this puzzle has therefore been rigorously proven to depend upon the assumption of how the total chime length is determined.
For start of chime 1, to END of chime ‘n’, the answer is NEGATIVE.
For start of chime 1, to START of chime ‘n’, the answer is POSITIVE!*
June 15, 2009 at 9:39 am |
For scenario a), the equation is:
Tt=n*Tc+(n-1)*Ts
(e.g.: #_#_#_#_#_#_#, for n=7)
For scenario b), the equation is:
Tt=n*Tc+(n-0)*Ts
=n*Tc+n*Ts
=n*(Tc+Ts)
(e.g.: #_#_#_#_#_#_, for n=7)
__________________
Lets work the a) case for both 7 & 10:
n=7?
a) Tt=n*Tc+(n-1)*Ts
=7*Tc+6*Ts
n=10?
a) Tt=n*Tc+(n-1)*Ts
=10*Tc+9*Ts
—————
Now the b) case for both 7 & 10:
n=7?
b) Tt=n*(Tc+Ts)
=7*(Tc+Ts)
n=10?
b) Tt=n*(Tc+Ts)
=10*(Tc+Ts)
June 15, 2009 at 9:40 am |
…For scenario a), the equation is:
Tt=n*Tc+(n-1)*Ts
(e.g.: #_#_#_#_#_#_#, for n=7)
For scenario b), the equation is:
Tt=n*Tc+(n-0)*Ts
=n*Tc+n*Ts
=n*(Tc+Ts)
(e.g.: #_#_#_#_#_#_, for n=7)
Lets work the a) case for both 7 & 10:
n=7?
a) Tt=n*Tc+(n-1)*Ts
=7*Tc+6*Ts
n=10?
a) Tt=n*Tc+(n-1)*Ts
=10*Tc+9*Ts
—————
Now the b) case for both 7 & 10:
n=7?
b) Tt=n*(Tc+Ts)
=7*(Tc+Ts)
n=10?
b) Tt=n*(Tc+Ts)
=10*(Tc+Ts)
June 15, 2009 at 9:41 am |
…Now, the question is: are n=7 & n=10 produce and equal (Total Period)/n in either of these scenarios?
This amounts to asking does:
a1)
Tt(7)/n==Tt(10)/n??
or
(7*Tc+6*Ts)/7==(10*Tc+9*Ts)/10 ?
or
b1)
7*(Tc+Ts)/7==10*(Tc+Ts)/10 ?
(this obviously simplifies to: Tc+Ts==Tc+Ts, which is trivially true.)
—————–
a1) Is true, provided that the silence lasts for 0 seconds, otherwise false.
b1) Is ALWAYS true!
So, I was guardedly correct** (No) for scenario a), but quite wrong for scenario b)!!
The resolution of this puzzle has therefore been rigorously proven to depend upon the assumption of how the total chime length is determined.
For start of chime 1, to END of chime ‘n’, the answer is NEGATIVE.
For start of chime 1, to START of chime ‘n’, the answer is POSITIVE!*
June 15, 2009 at 9:42 am |
…OK: The tricky notions, regarding the question as implying the temporal distance between the 7 o’clock chimes, and the 10 o’clock chimes.
There are several options available here for consideration:
The gaps:
7am to 10am ~ 3h
7am to 10pm ~ 3+12=15h
7pm to 10pm ~ 3+12=15h
7pm to 10pm ~ 3h
Which resolve down to two distinct durations:
3h & 15h
Both of which are subject to the a) & b) condition of the judgement of the length of a “strike”.
These trick answers are bogus, in my opinion, but don’t tell Richard that I said that, or he may join the BCA!
But do inform him (for that is, I understand, the prime target of these puzzles), that all of the above machinations had occurred to me in my head, (not to commit in writing), in under say, a minute.
Possibly closer to 30 seconds than a full minute.
It is a burden to be a systems analyst with Asperger’s!
_____________
* That the chime durations are equal, and the silence durations are equal.
** In assuming that the silent gaps were not zero length.
June 15, 2009 at 9:43 am |
P.S. Where are my 100 points?
I accept luncheon vouchers and book tokens.
I’m not proud.
June 15, 2009 at 9:58 am |
Pleased to report that I do not have to eat any humble pie (7/6*9=10.4999=~10.5). This reminds me of the kind of maths I was taught for the 11-plus. That dates me, doesn’t it!
I did not get the 3 hrs answer however; loved that one.
June 15, 2009 at 10:09 am |
Dammitt!! My answer was way out but its still possible….let me try and explain my thinking:
When in the question it said it takes 7 seconds to strike 7 ‘o’clock i took this literally.
Meaning that the time was 7 secs to 7 ‘o’clock
Taking that time and applying it to the 10 ‘o’clock situ, it would still take 7 secs as it would be 7 secs to 10 ‘o’clock.
So on that working, my answer was ‘7 seconds’!
Am i alone here or can anyone else see my logic? :s
June 15, 2009 at 10:20 am |
Yeah, to be honest, although I can see the logic in all of the scenarios, there’s nowhere near enough information for an even vaguely accurate answer… some clocks don’t even chime the number of hours, just have a set sound or series of sounds – for example every hour or quarter-hour – so it’s equally plausible that the time this takes is seven seconds, no matter which hour.
June 15, 2009 at 10:39 am |
BTW, the tricky answer has 4 possible solutions, including from the ‘3 hours’ one:
1) 3 hours
2) 3 hours LESS the chime duration
3) 15 hours
4) 15 hours LESS the chime duration
Consider the AM/PM conundrum.
June 15, 2009 at 10:49 am |
got it right!
June 15, 2009 at 11:27 am |
Am I seriously the only person who did what I did? Come on. Someone else has to have added the number of chimes to pauses and divided the time by it. I swear to god, it makes sense even if it IS wrong.
June 15, 2009 at 11:53 am |
@ScreamingGreenConure
As Ben Goldacre is fond of saying:
“It is not that simple!”
But you were correct, given a number of plausible assumptions…
June 15, 2009 at 12:01 pm |
The correct solution doesn’t take chime length into account, right? I don’t get it. I know I can’t know for certain that the chimes and pauses are exactly equal in length, but I know that the chime has to take longer than no time at all. If 7 o’clock has 7 chimes and 6 pauses, then the pauses can’t take 7/6 seconds, because then there would be no chimes. One chime plus one pause combined must take a little less than a second for 7 o’clock to take 7 seconds exactly.
My head hurts.
June 15, 2009 at 3:58 pm |
@Michael Gray: I think you made it more complicated than it needs to be. You don’t need to know the length of silence in between each chime. We just need to know the rate at which each new chime starts.
So in scenario a (Length of time from beginning to the end of the last chime) we just need to know how long the last chime goes for. and in scenario b we don’t even need to know that much.
Alex Garner did bring up the Westminster Quarters issue which would also have a large effect on the calculation.
In the end, the equation I came up with is as follows:
[(7-C-W)/6] * (H-1) + C + W
C: Length of last chime — This can be set to 0 for scenario a
W: Length of tune clock plays before hour chimes.
H: Hour of day.
So for the basic reading of how the question was intended the result is:
[(7-0-0)/6] * (10-1) + 0 + 0
(7/6) * 9 = 10.5
If we are counting the length of the last chime, then any combination for C & W where C + W = 1s would result in the clock taking 10s to chime for 10 o’clock.
June 15, 2009 at 5:15 pm |
Michael you have too much time on your hands!
June 15, 2009 at 7:19 pm |
I was thinking 3 hours and change, because he had to wait for ten o’clock to come.
June 16, 2009 at 12:39 am |
@Garrett
“…I think you made it more complicated than it needs to be. You don’t need to know the length of silence in between each chime…”
Not true for my scenario a), but true for scenario b).
See my references labelled a1) & b1)
June 16, 2009 at 9:48 am |
If I have 2 much time on my hands, how long would 7 much time be?
June 16, 2009 at 2:59 pm |
Just my two cents (or pence, depending on which side of the pond you’re on):
“Assuming” a little foolishness, and deciding that we are watching the clock throughout its cycles, it is interesting to note that as Americans we enjoy the perception that the hour is struck at first ‘chime’ of the hour. It is my understanding that Brits enjoy the perception that the hour is struck at the last ‘chime’ of the hour. Again, assuming we are discussing an “a.m. to a.m.” cycle:
In America, the cycle would be 3 hours.
In Great Britian, the cycle would be 3 hours and 3 seconds.
June 17, 2009 at 12:25 am |
[...] Answer to the Friday puzzle! Last week I posted this puzzle… Albert works in a clock factory. He has noticed that his favourite clock in the [...] [...]
June 23, 2009 at 11:29 am |
My initial response was to try to work out the timing between the strikes, but then the answer “3 hours and 7 seconds” occurred to me, and I liked that answer much better.
If it takes 7 seconds to strike 7 O’clock, then it must be 7 seconds to 7. It is therefore 3 hours and 7 seconds to 10 O’clock.
October 14, 2009 at 3:52 am |
This puzzle reminds me of a similar one: There are two riflemen. One can get off 5 shots in five seconds. The other can get off 10 shots in 10 seconds. Which rifleman is the faster shot? The time the shots themselves take is not a consideration.